/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A diffraction pattern forms when... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 25

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 675 nm. Determine the angle that locates the first dark fringe when the width of the slit is (a) \(1.8 \times 10^{-4} \mathrm{m}\) and \(\quad(\mathbf{b}) 1.8 \times 10^{-6} \mathrm{m}\)

Short Answer

Expert verified
For (a) \( \theta \approx 0.215 \text{ degrees} \); for (b) \( \theta \approx 22.02 \text{ degrees} \).

Step by step solution

01

Understand the Problem

The problem asks to find the angle that corresponds to the first dark fringe in a diffraction pattern. This occurs when light passes through a single slit and is related to the slit width and light wavelength.
02

Apply the Formula for Single-Slit Diffraction

The formula for locating dark fringes in single-slit diffraction is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of the dark fringe, \( m \) is the order of the dark fringe (1 for the first dark fringe), and \( \lambda \) is the wavelength of light.
03

Calculate the Angle for Part (a)

For part (a), where the slit width \( a = 1.8 \times 10^{-4} \: \mathrm{m} \) and \( \lambda = 675 \: \mathrm{nm} = 675 \times 10^{-9} \: \mathrm{m} \), we use the formula \( \sin \theta = \frac{m \lambda}{a} = \frac{1 \times 675 \times 10^{-9}}{1.8 \times 10^{-4}} \). Calculating this gives \( \sin \theta = 0.00375 \). Therefore, \( \theta = \arcsin(0.00375) \approx 0.215 \) degrees.
04

Calculate the Angle for Part (b)

For part (b), where the slit width \( a = 1.8 \times 10^{-6} \: \mathrm{m} \), we again apply the formula \( \sin \theta = \frac{1 \times 675 \times 10^{-9}}{1.8 \times 10^{-6}} \). Calculating this gives \( \sin \theta = 0.375 \). Therefore, \( \theta = \arcsin(0.375) \approx 22.02 \) degrees.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffraction Pattern
When light shines through a narrow opening like a single slit, it doesn't just pass straight through. Instead, the light spreads out and forms a pattern of alternating dark and bright regions. This phenomenon is known as a diffraction pattern. This pattern demonstrates the wave nature of light. Instead of moving in straight lines, the light waves spread out, interfere, and create areas with no light (dark fringes) and areas with intense light (bright fringes). The spacing and intensity of these fringes provide valuable insights into the properties of the light and the slit itself.
Wavelength of Light
The wavelength of light is a key factor in determining the characteristics of a diffraction pattern. Wavelength ( \( \lambda \) ) is the distance between successive peaks of a light wave. For this exercise, we are considering light with a wavelength of 675 nanometers (nm), which is part of the visible spectrum.
  • Wavelength affects how much the light will spread out after passing through the slit.
  • Longer wavelengths lead to wider spread and thus, wider diffraction patterns.
  • Shorter wavelengths cause a tighter, more compact pattern.
Knowing the wavelength helps you calculate the position of dark and bright fringes in the pattern.
Angle of Dark Fringe
In the diffraction pattern, dark fringes occur at specific angles where the light waves cancel each other out due to destructive interference. The angle of a dark fringe, especially the first one, is what this exercise focuses on. To find this angle, the relationship for single-slit diffraction is used: \[ a \sin \theta = m \lambda \] Where:
  • \( a \) is the slit width,
  • \( \theta \) is the angle of the dark fringe,
  • \( m \) is the order number of the fringe (1 for the first dark fringe),
  • \( \lambda \) is the wavelength of light.
This equation allows us to calculate the precise angle where the first dark fringe will appear, by solving for \( \theta \) using the known values of \( a \) and \( \lambda \).
Slit Width
The slit width ( \( a \) ) plays a crucial role in shaping the diffraction pattern. In the given problem, the widths provided are \(1.8 \times 10^{-4} \mathrm{m}\) and \(1.8 \times 10^{-6} \mathrm{m}\).
  • A larger slit width will cause the diffraction pattern to be narrower, with less spread between the fringes.
  • A smaller slit width increases the divergence of light, leading to a wider spread in the pattern.
  • The size of the slit also directly determines where the dark and bright fringes appear. The equation \( a \sin \theta = m \lambda \) directly relates slit width to the angle of diffraction.
Understanding the impact of slit width helps in predicting how the diffraction pattern will look under different conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A circular drop of oil lies on a smooth, horizontal surface. The drop is thickest in the center and tapers to zero thickness at the edge. When illuminated from above by blue light \((\lambda=455 \mathrm{nm}), 56\) concentric bright rings are visible, including a bright fringe at the edge of the drop. In addition, there is a bright spot in the center of the drop. When the drop is illuminated from above by red light \((\lambda=637 \mathrm{nm})\) a bright spot again appears at the center, along with a different number of bright rings. Ignoring the bright spot, how many bright rings appear in red light? Assume that the index of refraction of the oil is the same for both wavelengths.

In a Young's double-slit experiment the separation \(y\) between the second- order bright fringe and the central bright fringe on a flat screen is \(0.0180 \mathrm{~m}\) when the light has a wavelength of \(425 \mathrm{nm} .\) Assume that the angles that locate the fringes on the screen are small enough so that \(\sin \theta \approx \tan \theta .\) Find the separation \(y\) when the light has a wavelength of \(585 \mathrm{nm}\).

A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{m} .\) When light with a wavelength of \(\lambda_{1}=510 \mathrm{nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2} ),\) and a wavelength of \(\lambda_{2}=740 \mathrm{nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2} .\)

A nonreflective coating of magnesium fluoride \((n=1.38)\) covers the glass \((n=1.52)\) of a camera lens. Assuming that the coating prevents reflection of yellow-green light (wavelength in vacuum \(=565 \mathrm{nm}\) ), determine the minimum nonzero thickness that the coating can have.

There are 5620 lines per centimeter in a grating that is used with light whose wavelength is 471 nm. A flat observation screen is located at a distance of 0.750 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fluids

Read Explanation

Kinematics Physics

Read Explanation

Fields in Physics

Read Explanation

Modern Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.